Question

. Find two consecutive numbers whose squares have the sum 85.​

Answer 1

Answer:

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Step-by-step explanation:

Let the numbers be x and (x+1) respectively,

                           ATQ,

  x²+(x+1)²=85

⇒x²+x²+1²+2(x)(1)=85                  {after expanding (x+1)²}

⇒  2x²+1+2x=85

⇒2x²+2x=85-1

⇒2x²+2x=84

⇒2(x²+x)=84

⇒x²+x=$\frac{84}{2}$

⇒x²+x=42

⇒x²+x-42=0

⇒x²+7x-6x-42=0                                      {middle-term splitting}

⇒x(x+7)-6(x+7)=0

⇒(x-6)(x+7)=0

When (x+7)=0 then x=6 {leave the other value because square can never              

                                                           be negative}    

Hence, the consecutive numbers = 6 and (x+1)=(6+1)=7